/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q58P A horizontal platform in the sha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the centre of the disk. The platform has a mass of 150kg, a radius of2.0m, and a rotational inertia of300kg.m2about the axis of rotation. A60kgstudent walks slowly from the rim of the platform toward the centre. If the angular speed of the system is1.5rad/swhen the student starts at the rim, what is the angular speed when she is0.50mfrom the centre?

Short Answer

Expert verified

Angular speed of student when she is at 0.5mfrom the center of a circular disk isÓ¬f=2.6rads2.

Step by step solution

01

Step 1: Given

  1. Mass of the platform,m=150kg
  2. Radius of platform,r=2.0m
  3. Initial angular speed of the systemÓ¬i=1.5rad/s
  4. Distance of the student from the center,R=0.5m
  5. Idisk=300kg.m2
02

Determining the concept

Calculate initial and final rotational inertias. Then, apply law of conservation of angular momentum to find final angular speed. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

Istudent=Mstudent×R2

Initialangularmomentum=Finalangularmomentum

Where,Istudent is moment of inertia of student, Mstudentis mass of student and R is radius.

03

Determining the angular speed of student when she is at 0.5 m from the center of a circular disk 

The initial rotational inertia of the system,

I=Idisk+Istudent

I=300+60×22=540kg.m2

The final rotational inertia when she reaches0.5mfrom the center,

If=Idisk+IstudentIf=300+60×0.52=315kg.m2

According to law of conservation of angular momentum:

Li=LfIiӬi=IfӬf540×1.5=315×Ӭf

Hence,

Ó¬f=2.6rads2

Hence,angular speed of student when she is at0.5mfrom the center of a circular disk isÓ¬f=2.6rads2.

Therefore, using law of conservation of angular momentum, the final angular velocity of the student at the given distance can be found.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform block of granite in the shape of a book has face dimensions of 20 cm and 15 cmand a thickness of 1.2 cm. The density (mass per unit volume) of granite is2.64 g/cm3. The block rotates around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that axis is 0.104 kg.m2/s. What is its rotational kinetic energy about that axis?

A car travels at80 km/hon a level road in the positive direction of an xaxis. Each tire has a diameter of66 c³¾.Relative to a woman riding in the car, and in unit-vector notation,whatis the velocityv→at the (a) center (b) top, and (c) bottom of the tire? What is the magnitude aof the acceleration at the (d) center (e)top, and (f)the bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what is the velocityv→at the (g) centre (h) top, and (i) bottom of the tire? And the magnitude aof the acceleration at the (j) centre (k) top(l) bottom of each tire?

In Figure, two skaters, each of mass 50kg, approach each other along parallel paths separated by3.0m. They have opposite velocities ofeach 1.4m/s. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the centre of the pole. Assume that the friction between skates and ice is negligible.

(a) What is the radius of the circle?

(b) What are the angular speeds of the skaters?

(c) What is the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by1.0m.

(d) What then are their angular speed?

(e) What then are the kinetic energy of the system?

(f) What provided the energy for the increased kinetic energy?

Figure is an overhead view of a thin uniform rod of length 0.600mand mass Mrotating horizontally at 80.0rad/scounter clock-wise about an axis through its centre. A particle of mass M/3.00and travelling horizontally at speed 40.0m/shits the rod and sticks. The particle’s path is perpendicular to the rod at the instant of the hit, at a distance dfrom the rod’s centre.

(a) At what value of dare rod and particle stationary after the hit?

(b) In which direction do rod and particle rotate if dis greater than this value?

Force F⇶Ä=(2.0N)i^-(3.0N)k^acts on a pebble with position vectorr⇶Ä=(0.50m)j^-(2.0m)k^relative to the origin. In unit- vector notation, (a) What is the resulting torque on the pebble about the origin and (b) What is the resulting torque on the pebble about the point(2.0m,0,-3.0m)?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.